Mathematics of Physics and Engineering

Velocity and Acceleration 35

and v • r — 0. So v is tangent to the circle. The acceleration a is

a(t) =v'(t) = - R(9"(t) sin 9(t) - (9'(t))^2 cos<?(*)) I

+ R(6"(t) cos8(t) - {9'(t))^2 sm9(t))j

or

a=-(0')^2 r+(6"/6')v. (1.3.21)

Thus, the tangential component of a is (9"/6')v, and the normal com-
ponent, also known as the centripetal acceleration, is ~{9')^2 r. Also,

H| = zV(0')^4 + (0")^2 -

EXERCISE 1.3.13.B Verify that (1.3.21) coincides with (1.3.19). Hint: First
verify that ds/dt = R6'{t) and du(t)/dt = -(0'(t)/R)r(t).

If the rotation is uniform with constant angular speed LJ, then 0(t) = cot
and we have the familiar expressions ||a|| = u>^2 R = ||i>||^2 /.R.
Note that the centripetal acceleration is in the direction of — r, that is,
in the direction toward the center. It is not a coincidence that the Latin
verb petere means "to look for."
Next, we write the decomposition (1.3.19) for the GENERAL PLANAR
MOTION IN POLAR COORDINATES (r, 9). Consider a frame with origin O
and fixed cartesian coordinate system (i, j, k) so that the motion is in the
(i, j) plane. Recall that, for a point P with position vector r, r = \r\, and
9 is the angle from vector i tor. Let r = r/r be the unit radius vector
and let 9 be the unit vector orthogonal to r so that r x 9 = i x j; draw a
picture or see Figure 2.1.3 on page 48 below. Then

{

r = cos#£ + sin0j,

8 = - sin 9 i + cos 9 j.

The vectors f, 9 are functions of 9. From (1.3.22) we get

{

dr/d9 = — sin#z + cos9j= 0,

d9/d9 = - cos9i — sin# j = —f.

(1.3.22)

(1.3.23)

Let r(t) be the position of the point mass m at time t. In polar
coordinates, r(t) = r(t)r(9(t)). The velocity of m in the frame O is
v — dr/dt = d(r(t)r(8(t)))/dt. Using the rule (1.3.4) and the chain rule,
we get v = (dr/dt)f+ r (dr/d9)(d9/dt), or

v = fr + r90 = fr + nJ0. (1.3.24)